Problem: Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{-10y^3 + 130y^2 - 420y}{4y^2 - 28y + 24}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ z = \dfrac {-10y(y^2 - 13y + 42)} {4(y^2 - 7y + 6)} $ $ z = -\dfrac{10y}{4} \cdot \dfrac{y^2 - 13y + 42}{y^2 - 7y + 6} $ Simplify: $ z = - \dfrac{5y}{2} \cdot \dfrac{y^2 - 13y + 42}{y^2 - 7y + 6}$ Next factor the numerator and denominator. $ z = - \dfrac{5y}{2} \cdot \dfrac{(y - 6)(y - 7)}{(y - 6)(y - 1)}$ Assuming $y \neq 6$ , we can cancel the $y - 6$ $ z = - \dfrac{5y}{2} \cdot \dfrac{y - 7}{y - 1}$ Therefore: $ z = \dfrac{ -5y(y - 7)}{ 2(y - 1)}$, $y \neq 6$